Why does a normal ordered product of operators (in CFT) have 0 expectation value?

Question

Why does a normal ordered product of operators (in CFT) have 0 expectation value?

The definition (Francesco - Conformal field theory pg. 174) of the normal ordered product of two operators $A(z)$ $B(z)$ is: $$(AB)(w) = \frac{1}{2\pi i} \oint_w \frac{dz}{z-w}A(z)B(w).\tag{6.130}$$

The definition of expectation value of operators $O_1(t_1)O_2(t_2)$ is $$\langle O_1(t_1)O_2(t_2)\rangle = \langle 0|T(O_1(t_1)O_2(t_2))|0 \rangle ,$$ i.e. the time ordered product sandwiched between vacuum states.

How to see from these definitions that we would have: $$ \langle(AB)(w)\rangle = 0?$$

Someone suggested to me that one should do a mode expansion. This is done in (Francesco - Conformal field theory pg. 175), it gives

$$(AB)(z) = \sum\limits_n z^{-n-h_{A}-h_{B}}(AB)_n,\tag{6.145}$$ where $$(AB)_m = \sum\limits_{n \le -h_A}A_n B_{m-n} + \sum\limits_{n > -h_A} B_{m-n}A_n.\tag{6.144}$$

If we could somehow interpret the modes $A_n$, $B_n$ as annihilation of creation operators, than that might be a way to get to 0 expectation value. But it is not clear to me how to get such an interpretation (for arbitrary operators $A$ and $B$).

Answer 1

For an operator with scaling dimension $\Delta \neq 0$ in a $d$-dimensional CFT, $\left < O(x) \right > = 0$ because there is no function of $x$ alone which has the right conformal covariance properties. A constant is Lorentz invariant but doesn't scale properly whereas $|x|^{-\Delta}$ scales properly but is not translation invariant. From this perspective, it doesn't matter whether $O(x)$ happens to be the normal ordered product of two other operators or not.

The approach you're trying works for a chiral operator of dimension $h$ in a 2d CFT if you use \begin{equation} O_n \left | 0 \right > = 0, \quad n > -h. \end{equation} You should be able to get this relation as a consequence of unitarity by using \begin{equation} [L_m, O_n] = [(h - 1)m - n] O_{m + n} \end{equation} which follows from the $T(z) O(0)$ OPE. But more generally, the modes with $n > -h$ have to be annihilation operators because \begin{equation} O(z) = \sum_n O_n z^{-n-h} \end{equation} and we need $O(0)$ to be able to act on the vacuum and not diverge. If something doesn't have this property, correlation functions of it are not defined and therefore we shouldn't call it an operator in a CFT.